Question

If roots of the equation $(a-b)x^2+(c-a)x+(b-c)=0$ are equal, then find progression in which $a,b,c$ lie.

Answer 1

$\Rightarrow$ The given equation $(a-b)x^2+(c-a)x+(b--c)=0$, comparing it with $A{x}^2+Bx+C=0$

Here, $A=a-b,B=c-a,C=b-c$

It is given that roots are equal.

$\therefore$ $Discriminant=0$

$\Rightarrow$ $B^2-4AC=0$

$\Rightarrow$ $(c-a{)}^2-4(a-b\left)\right(b-c)=0$

$\Rightarrow$ $c^2+a^2-2ac-4(ab-ac-b^2+bc)$

$\Rightarrow$ $c^2+a^2-2ac-4ab+4ac+4b^2-4bc$

$\Rightarrow$ $c^2+a^2+2ac-4b(a+c)+4b^2$

$\Rightarrow$ $(a+c{)}^2-4b(a+c)+4b^2$

$\Rightarrow$ $\left[\right(a+c)-2b{]}^2=0$

$\Rightarrow$ $(a+c)-2b=0$

$\Rightarrow$ $a+c=2b$

$\therefore$ $b=\frac{a+c}{2}$

$\therefore$ $a,b,c$ are in $A.P.$